The coefficient of y in the expansion of `(y^(2) + c//y)^(5)` is

Options

A. 29c
B. 10 c
C. `10c^(3)`
D. `20 c^(2)`

Answer

To find the coefficient of \( y \) in the expansion of \( (y^2 + \frac{c}{y})^5 \), we will use the Binomial Theorem.

### Step-by-step Solution:

1. **Identify the terms in the binomial**:
We have \( p = y^2 \) and \( q = \frac{c}{y} \) with \( n = 5 \).

2. **Write the general term**:
The general term \( T_{r+1} \) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
Substituting the values, we get:
\[ T_{r+1} = \binom{5}{r} (y^2)^{5-r} \left(\frac{c}{y}\right)^r \]

3. **Simplify the general term**:
This becomes:
\[ T_{r+1} = \binom{5}{r} (y^{2(5-r)}) \left(\frac{c^r}{y^r}\right) = \binom{5}{r} c^r y^{10 - 3r} \]

4. **Find the coefficient of \( y \)**:
We need to find the value of \( r \) such that the exponent of \( y \) is 1:
\[ 10 - 3r = 1 \]
Solving for \( r \):
\[ 10 - 1 = 3r \implies 9 = 3r \implies r = 3 \]

5. **Substitute \( r \) back into the general term**:
Now substitute \( r = 3 \) into the general term to find the coefficient:
\[ T_{4} = \binom{5}{3} c^3 y^{10 - 3 \cdot 3} \]
Simplifying this gives:
\[ T_{4} = \binom{5}{3} c^3 y^{1} \]

6. **Calculate \( \binom{5}{3} \)**:
We know that:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

7. **Final coefficient**:
Therefore, the coefficient of \( y \) is:
\[ 10c^3 \]

### Conclusion:
The coefficient of \( y \) in the expansion of \( (y^2 + \frac{c}{y})^5 \) is \( 10c^3 \).

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